Question: What is the value of $\dfrac{d}{dx}(x^3-4x^2+3x)$ at $x=5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $38$ (Choice B) B $8$ (Choice C) C $40$ (Choice D) D $68$
Let's first find the expression for $\dfrac{d}{dx}(x^3-4x^2+3x)$ and then evaluate it at $x=5$. According to the sum rule, the derivative of $x^3-4x^2+3x$ is the sum of the derivatives of $x^3$, $-4x^2$, and $3x$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\dfrac{d}{dx}(x^3)=3x^2$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(x^3-4x^2+3x) \\\\ &=\dfrac{d}{dx}(x^3)-4\dfrac{d}{dx}(x^2)+3\dfrac{d}{dx}(x)&&\gray{\text{Basic differentiation rules}} \\\\ &=3x^2-4\cdot2x+3\cdot 1x^0&&\gray{\text{The power rule}} \\\\ &=3x^2-8x+3 \end{aligned}$ So we found that $\dfrac{d}{dx}(x^3-4x^2+3x)=3x^2-8x+3$. Plugging in $x=5$ and evaluating using the calculator, we find that the value is $38$. In conclusion, the value of $\dfrac{d}{dx}(x^3-4x^2+3x)$ at $x=5$ is $38$.